Skip to content
Related Articles

Related Articles

Change the given string according to the given conditions
  • Last Updated : 15 Oct, 2020

Given a string S, the task is to change the string id it doesn’t follow any of the rules given below and print the updated string. The rules for the proofreading are: 

  1. If there are three consecutive characters, then it’s a wrong spell. Remove one of the characters. For Example string “ooops” can be changed to “oops”.
  2. If two pairs of the same characters (AABB) are connected together, it’s a wrong spell. Delete one of the characters of the second pair. For Example string “helloo” can be changed to “hello”.
  3. The rules follow the priority from left to right.

Examples: 

Input: S = “helloo”
Output: hello
Explanation:
As per the Rule #2
helloo => hello

Input: S = “woooow”
Output: woow
Explanation: 
As per the Rule #2
woooow => wooow
As per the Rule #1
wooow => woow

Approach: The idea is to traverse the string and if there is a wrong spelling, remove the extra characters according to the given conditions. As the priority of errors is from left to right, and according to the rules given, it can be seen that the judgment of spelling errors will not conflict. Consider traversing from left to right, adding the already legal characters to the result. Below are the steps:



  • Initialize a stack to store the characters and to compare the last characters of the string.
  • Traverse the string and add the character to the stack.
  • Check the last 3 characters of the stack, if the same then pop the character at the top of the stack.
  • Check the last 4 characters of the stack, if the same then pop the character at the top of the stack.
  • Finally, return the characters of the stack.

Below is the implementation of the above approach:
 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to proofread the spells
string proofreadSpell(string& str)
{
    vector<char> result;
 
    // Loop to iterate over the
    // characters of the string
    for (char c : str) {
 
        // Push the current character c
        // in the stack
        result.push_back(c);
 
        int n = result.size();
 
        // Check for Rule 1
        if (n >= 3) {
            if (result[n - 1]
                    == result[n - 2]
                && result[n - 1]
                       == result[n - 3]) {
                result.pop_back();
            }
        }
        n = result.size();
 
        // Check for Rule 2
        if (n >= 4) {
            if (result[n - 1]
                    == result[n - 2]
                && result[n - 3]
                       == result[n - 4]) {
                result.pop_back();
            }
        }
    }
 
    // To store the resultant string
    string resultStr = "";
 
    // Loop to iterate over the
    // characters of stack
    for (char c : result) {
        resultStr += c;
    }
 
    // Return the resultant string
    return resultStr;
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "helloo";
 
    // Function Call
    cout << proofreadSpell(str);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to proofread the spells
public static String proofreadSpell(String str)
{
    Vector<Character> result = new Vector<Character>();
 
    // Loop to iterate over the
    // characters of the string
    for(int i = 0; i < str.length(); i++)
    {
         
        // Push the current character c
        // in the stack
        result.add(str.charAt(i));
 
        int n = result.size();
 
        // Check for Rule 1
        if (n >= 3)
        {
            if (result.get(n - 1) ==
                result.get(n - 2) &&
                result.get(n - 1) ==
                result.get(n - 3))
            {
                result.remove(result.size() - 1);
            }
        }
        n = result.size();
 
        // Check for Rule 2
        if (n >= 4)
        {
            if (result.get(n - 1) ==
                result.get(n - 2) &&
                result.get(n - 3) ==
                result.get(n - 4))
            {
                result.remove(result.size() - 1);
            }
        }
    }
 
    // To store the resultant string
    String resultStr = "";
 
    // Loop to iterate over the
    // characters of stack
    for(Character c : result)
    {
        resultStr += c;
    }
     
    // Return the resultant string
    return resultStr;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given string str
    String str = "helloo";
 
    // Function call
    System.out.println(proofreadSpell(str));
}
}
 
// This code is contributed by divyeshrabadiya07

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for
# the above approach
 
# Function to proofread
# the spells
def proofreadSpell(str):
   
    result = []
 
    # Loop to iterate over the
    # characters of the string
    for c in str:
 
        # Push the current character c
        # in the stack
        result.append(c)
        n = len(result)
 
        # Check for Rule 1
        if(n >= 3):
            if(result[n - 1] == result[n - 2] and
               result[n - 1] == result[n - 3]):
                result.pop()
        n = len(result)
 
        # Check for Rule 2
        if(n >= 4):
            if(result[n - 1] == result[n - 2] and
               result[n - 3] == result[n - 4]):
                result.pop()
 
    # To store the
    # resultant string
    resultStr = ""
 
    # Loop to iterate over the
    # characters of stack
    for c in result:
        resultStr += c
 
    # Return the resultant string
    return resultStr
 
# Driver Code
 
# Given string str
str = "helloo"
 
# Function Call
print(proofreadSpell(str))
 
# This code is contributed by avanitrachhadiya2155

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach 
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
// Function to proofread the spells
static string proofreadSpell(string str)
{
     
    // ArrayList result=new ArrayList();
    List<char> result = new List<char>();
     
    // Loop to iterate over the
    // characters of the string
    foreach(char c in str)
    {
 
        // Push the current character c
        // in the stack
        result.Add(c);
 
        int n = result.Count;
 
        // Check for Rule 1
        if (n >= 3)
        {
            if (result[n - 1] == result[n - 2] &&
                result[n - 1] == result[n - 3])
            {
                result.RemoveAt(n - 1);
            }
        }
         
        n = result.Count;
 
        // Check for Rule 2
        if (n >= 4)
        {
            if (result[n - 1] == result[n - 2] &&
                result[n - 3] == result[n - 4])
            {
                result.RemoveAt(n - 1);
            }
        }
    }
 
    // To store the resultant string
    string resultStr = "";
 
    // Loop to iterate over the
    // characters of stack
    foreach(char c in result)
    {
        resultStr += c;
    }
 
    // Return the resultant string
    return resultStr;
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Given string str
    string str = "helloo";
 
    // Function call
    Console.Write(proofreadSpell(str));
}
}
 
// This code is contributed by rutvik_56

chevron_right


Output: 

hello

 

Time Complexity: O(N) 
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :